ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 346

\[1)\ f(x) = x + \sqrt{x^{2} - 1};\]

\[f^{'}(x) = 1 + \frac{2x}{2\sqrt{x^{2} - 1}} =\]

\[= \frac{\sqrt{x^{2} - 1} + x}{\sqrt{x^{2} - 1}};\]

\[= \frac{\left( \sqrt{x^{2} - 1} + x \right)\left( \sqrt{x^{2} - 1} - x \right)}{\left( x^{2} - 1 \right)\sqrt{x^{2} - 1}} =\]

\[= \frac{x^{2} - 1 - x^{2}}{\left( x^{2} - 1 \right)\sqrt{x^{2} - 1}} =\]

\[= \frac{- 1}{\left( x^{2} - 1 \right)\sqrt{x^{2} - 1}} < 0.\]

\[Промежуток\ возрастания:\]

\[\sqrt{x^{2} - 1} + x \geq 0\]

\[\sqrt{x^{2} - 1} \geq - x\]

\[x^{2} - 1 \geq x^{2}\]

\[- x \leq 0\]

\[x \geq 0.\]

\[Область\ определения:\]

\[x^{2} - 1 \geq 0\]

\[(x + 1)(x - 1) \geq 0\]

\[x \leq - 1;\text{\ \ \ x} \geq 1.\]

\[\lim_{x \rightarrow \text{-}\infty}\frac{f(x)}{x} = \lim_{x \rightarrow - \infty}\left( x + \sqrt{x^{2} - 1} \right) =\]

\[= \lim_{x \rightarrow - \infty}\frac{x^{2} - x^{2} + 1}{x - \sqrt{x^{2} - 1}} = \frac{1}{\infty} = 0;\]

\[k = \lim_{x \rightarrow + \infty}\frac{f(x)}{x} =\]

\[\lim_{x \rightarrow + \infty}\left( 1 + \frac{\sqrt{x^{2} - 1}}{x} \right) = 1 + \frac{1}{1} = 2;\]

\[b = \lim_{x \rightarrow + \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow + \infty}\left( x + \sqrt{x^{2} - 1} - 2x \right) =\]

\[= \lim_{x \rightarrow + \infty}\left( \sqrt{x^{2} - 1} - x \right) =\]

\[= \lim_{x \rightarrow + \infty}\frac{x^{2} - 1 - x^{2}}{\sqrt{x^{2} - 1} + x} = - \frac{1}{\infty} = 0.\]

\[y = 0;\ \ \ y = 2x.\]

\[2)\ f(x) = x - \sqrt{x^{2} - 2x};\]

\[f^{'}(x) = 1 - \frac{2x - 2}{2\sqrt{x^{2} - 2x}} =\]

\[= \frac{\sqrt{x^{2} - 2x} - x + 1}{\sqrt{x^{2} - 2x}};\]

\[= \frac{(x - 1)(x - 1)}{\left( x^{2} - 2x \right)\sqrt{x^{2} - 2x}} =\]

\[= \frac{(x - 1)^{2}}{\left( x^{2} - 2x \right)\sqrt{x^{2} - 2x}} > 0.\]

\[Промежуток\ возрастания:\]

\[\sqrt{x^{2} - 2x} - x + 1 \geq 0\]

\[\sqrt{x^{2} - 2x} \geq x - 1\]

\[x^{2} - 2x \geq x^{2} - 2x + 1\]

\[x - 1 \leq 0\]

\[x \leq 1.\]

\[Область\ определения:\]

\[x^{2} - 2x \geq 0\]

\[x(x - 2) \geq 0\]

\[x \leq 0;\text{\ \ \ x} \geq 2.\]

\[\lim_{x \rightarrow + \infty}\frac{f(x)}{x} = \lim_{x \rightarrow + \infty}\left( x - \sqrt{x^{2} - 2x} \right) =\]

\[= \lim_{x \rightarrow + \infty}\frac{x^{2} - x^{2} + 2x}{x + \sqrt{x^{2} - 2x}} = \frac{2}{1 + 1} = 1;\]

\[k = \lim_{x \rightarrow - \infty}\frac{f(x)}{x} =\]

\[= \lim_{x \rightarrow - \infty}\left( 1 - \frac{\sqrt{x^{2} - 2x}}{x} \right) =\]

\[= 1 + \frac{1}{1} = 2;\]

\[b = \lim_{x \rightarrow - \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow - \infty}\left( x - \sqrt{x^{2} - 2x} - 2x \right) =\]

\[= \lim_{x \rightarrow - \infty}\left( \sqrt{x^{2} - 2x} - x \right) =\]

\[= \lim_{x \rightarrow - \infty}\frac{x^{2} - 2x - x^{2}}{\sqrt{x^{2} - 2x} + x} =\]

\[= \frac{- 2}{1 + 1} = - 1;\]

\[y = 1;\ \ \ y = 2x - 1.\]

\[3)\ f(x) = x^{2} \bullet e^{- x};\]

\[f^{'}(x) = 2x \bullet e^{- x} - x^{2} \bullet e^{- x} =\]

\[= e^{- x}\left( 2x - x^{2} \right);\]

\[f^{''}(x) = - e^{- x}\left( 2x - x^{2} \right) + e^{- x}(2 - 2x) =\]

\[= e^{- x}\left( 2 - 2x - 2x + x^{2} \right) =\]

\[= e^{- x}\left( x^{2} - 4x + 2 \right).\]

\[Промежуток\ возрастания:\]

\[2x - x^{2} \geq 0\]

\[x(x - 2) \leq 0\]

\[0 \leq x \leq 2.\]

\[Выпукла\ вниз:\]

\[x^{2} - 4x + 2 \geq 0\]

\[D = 16 - 8 = 8\]

\[x = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2};\]

\[x \leq 2 - \sqrt{2};\ \ \ x \geq 2 + \sqrt{2}.\]

\[\lim_{x \rightarrow + \infty}{f(x)} = \lim_{x \rightarrow + \infty}\left( x^{2} \bullet e^{- x} \right) =\]

\[= \lim_{x \rightarrow + \infty}\frac{x^{2}}{e^{x}} = 0;\]

\[y = 0.\]

\[4)\ f(x) = x^{3} \bullet e^{- x};\]

\[f^{'}(x) = 3x^{2} \bullet e^{- x} - x^{3} \bullet e^{- x} =\]

\[= e^{- x}\left( 3x^{2} - x^{3} \right);\]

\[f^{''}(x) =\]

\[= - e^{- x}\left( 3x^{2} - x^{3} \right) + e^{- x}\left( 6x - 3x^{2} \right) =\]

\[= e^{- x}\left( 6x - 3x^{2} - 3x^{2} + x^{3} \right) =\]

\[= e^{- x}\left( x^{3} - 6x^{2} + 6x \right).\]

\[Промежуток\ возрастания:\]

\[3x^{2} - x^{3} \geq 0\]

\[3 - x \geq 0\]

\[x \leq 3;\text{\ \ \ x} = 0.\]

\[Выпукла\ вниз:\]

\[x^{3} - 6x^{2} + 6x \geq 0\]

\[x\left( x^{2} - 6x + 6 \right) \geq 0\]

\[D = 36 - 24 = 12\]

\[x = \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 2\sqrt{3}}{2} = 3 \pm \sqrt{3};\]

\[0 \leq x \leq 3 - \sqrt{3};\ \ \ x \geq 3 + \sqrt{3}.\]

\[\lim_{x \rightarrow + \infty}{f(x)} = \lim_{x \rightarrow + \infty}\left( x^{3} \bullet e^{- x} \right) =\]

\[= \lim_{x \rightarrow + \infty}\frac{x^{3}}{e^{x}} = 0;\]

\[y = 0.\]

\[5)\ f(x) = \frac{(x - 1)^{3}}{x^{2}};\]

\[f^{'}(x) =\]

\[= \frac{3(x - 1)^{2} \bullet x^{2} - (x - 1)^{3} \bullet 2x}{x^{4}} =\]

\[= \frac{(x - 1)^{2} \bullet \left( 3x^{2} - 2x^{2} + 2x \right)}{x^{4}} =\]

\[= \frac{(x - 1)^{2} \bullet \left( x^{2} + 2x \right)}{x^{4}};\]

\[= \frac{4x^{7} - 6x^{5} + 2x^{4} - 4x^{7} + 12x^{5} - 8x^{4}}{x^{8}} =\]

\[= \frac{6x^{5} - 6x^{4}}{x^{8}}.\]

\[Промежуток\ возрастания:\]

\[(x - 1)^{2} \bullet \left( x^{2} + 2x \right) \geq 0\]

\[x(x + 2) \geq 0\]

\[x \leq - 2;\ \ \ x \geq 0;\text{\ \ \ x} = 1.\]

\[Выпукла\ вниз:\]

\[6x^{5} - 6x^{4} \geq 0\]

\[x - 1 \geq 0\]

\[x \geq 1;\]

\[x \neq 0.\]

\[k = \lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\frac{(x - 1)^{3}}{x^{3}} =\]

\[= \frac{1}{1} = 1;\]

\[b = \lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow \infty}\left( \frac{(x - 1)^{3}}{x^{2}} - x \right) =\]

\[= \lim_{x \rightarrow \infty}\frac{x^{3} - 3x^{2} + 3x - 1 - x^{3}}{x^{2}} =\]

\[= \frac{- 3 + 0 - 0}{1} = - 3;\]

\[x = 0;\ \ \ y = x - 3.\]

\[6)\ f(x) = \frac{x^{4}}{(x + 1)^{3}};\]

\[f^{'}(x) =\]

\[= \frac{4x^{3} \bullet (x + 1)^{3} - x^{4} \bullet 3(x + 1)^{2}}{(x + 1)^{6}} =\]

\[= \frac{(x + 1)^{2}\left( 4x^{4} + 4x^{3} - 3x^{4} \right)}{(x + 1)^{6}} =\]

\[= \frac{x^{4} + 4x^{3}}{(x + 1)^{4}};\]

\[= \frac{12x^{2}}{(x + 1)^{5}}.\]

\[Промежуток\ возрастания:\]

\[x^{4} + 4x^{3} \geq 0\]

\[x^{3}(x + 4) \geq 0\]

\[x \leq - 4;\text{\ \ \ x} \geq 0.\]

\[Выпукла\ вниз:\]

\[x + 1 \geq 0\]

\[x \geq - 1;\]

\[x \neq - 1.\]

\[k = \lim_{x \rightarrow \infty}\frac{f(x)}{x} = \lim_{x \rightarrow \infty}\frac{x^{3}}{(x + 1)^{3}} =\]

\[= \frac{1}{1} = 1;\]

\[b = \lim_{x \rightarrow \infty}\left( f(x) - kx \right) =\]

\[= \lim_{x \rightarrow \infty}\left( \frac{x^{4}}{(x + 1)^{3}} - x \right) =\]

\[= \lim_{x \rightarrow \infty}\frac{x^{4} - x^{4} - 3x^{3} - 3x^{2} - x}{x^{3} + 3x^{2} + 3x + 1} =\]

\[= \frac{- 3 - 0 - 0}{1 + 0 + 0 + 0} = - 3.\]

\[x = - 1;\ \ \ y = x - 3.\]